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Theorem: Every non-constant polynomial with complex coefficients has a zero in $\mathbb{C}$.

Proof: Let $P(z)$ be any non-constant polynomial. Assume by way of contradiction that $P(z)\neq 0$ for all $z\in\mathbb{C}$. Then the function $f(z):=1/P(z)$ is an entire function. Since $P$ is non-constant, $P\rightarrow \infty$ as $z\rightarrow\infty$, so that $f$ is bounded. By Liouville's Theorem $f$ is constant. Then $P$ is constant, a contradiction. Hence $P$ has a zero in $\mathbb{C}$.

I am trying to understand this proof, but I am stuck on "$P\rightarrow\infty$ as $z\rightarrow\infty$." Could somebody provide further details, please?

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    $\begingroup$ $P(z)=z^n+\ldots $, with leading term $z^n$. Show that this means $P\to \infty$ for $z\to \infty$. $\endgroup$ – Dietrich Burde Feb 23 '15 at 20:53
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Assume that $P(z) = a_nz^n + a_{n-1}z^{n-1} + \dots + a_1 z + a_0$. Then $P(z) = z^n \left( a_n + \frac{a_{n-1}}{z} + \dots + \frac{a_1}{z^{n-1}} + \frac{a_0}{z^n}\right)$, and so if $z\to \infty$, you might notice something happens with the other terms.

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